Residue Theorem

• Jul 9th 2009, 11:04 PM
Richmond
Residue Theorem
Calculate the following integral using Residue Theorem

$\int \frac{dz}{(1+z^2)^2}$

from $\infty$ to $-\infty$
• Jul 9th 2009, 11:42 PM
Moo
Hello,

Let $f(z)=\frac{1}{(1+z^2)^2}$

Obviously, i and -i are double poles, because $1+z^2=(z+i)(z-i)$

So, using the general formula for a pole of order n : $\text{Res}_a f(z)=\frac{1}{(n-1)!} \lim_{z\to a}\frac{\partial^{n-1}}{\partial z^{n-1}} (z-a)^n f(z)$
we easily get the residue of f at i (the one that will be useful for the computation) :
$\text{Res}_{z=i} f(z)=\lim_{z\to i} \frac{d}{dz} (z-i)^2 f(z)=\lim_{z\to i} \frac{d}{dz} \frac{1}{(z+i)^2}$
which is easy to compute.

Spoiler:
the residue is $-\frac i4$

Now, use the semicircle contour :

And keep the poles with positive imaginary part. So only i.

And then $\int_{-\infty}^\infty f(z) ~dz=2i\pi \text{Res}_{z=i} f(z)$
• Jul 10th 2009, 12:22 AM
Richmond
ok. so for

$Res(f,i) = \lim_{z\to i} \frac{d}{dz}(z-i)^2f(z)$
$= \lim_{z\to i} \frac{d}{dz} \frac {1}{z+i}^2$
$=\lim_{z\to i} \frac{-(2z+2i)}{(z+i)^4}$
$= \frac {-4i}{16i^4}
= \frac{i}{4}$

and

$Res(f,-i) = \lim_{z\to -i} \frac{d}{dz}(z+i)^2f(z)$
$= \lim_{z\to -i} \frac{d}{dz} \frac {1}{z-i}^2$
$=\lim_{z\to -i} \frac{-(2z-2i)}{(z-i)^4}$
$= \frac {4i}{16i^4}
= \frac {-i}{4}$

Am I right to say that you took semi contour and the positive side of the imaginary because of the -infinity to infinity?

The last part would be

$\int_{-\infty}^\infty f(z) ~dz=2\pi i Res(f,i)$
$= 2 \pi i * \frac {i}{4} = \frac {-\pi}{2}$

• Jul 10th 2009, 12:32 AM
Moo
Hey...

$i^4=(i^2)^2=(-1)^2=1$

So $= \frac {-4i}{16i^4}
= {\color{red}-}\frac{i}{4}
$

You got wrong for the second one too.

Isn't it surprising to get a negative answer for the integral of a positive function ?

Quote:

Am I right to say that you took semi contour and the positive side of the imaginary because of the -infinity to infinity?
Yes. If we had the integral from infinity to -infinity, it would have been more logic to take the semi circle and the negative side of the imaginary.
But the result is actually the same since $\int_{-\infty}^\infty =-\int_\infty^{-\infty}$

But more commonly, we take the upper semi circle ;)

I hope I'm clear :s
• Jul 10th 2009, 12:36 AM
Richmond
Ok. thanks for the clearance..

Yeap. I got a careless calculation error there, thanks for pointing that out.